Groovy中 - 加入过滤器映射

Question

我想查谁能够拿出最好的Groovy-SH的方式来实现这一目标 -

def m1 = [["id":"1","o":"11"],["id":"1","o":"12"],["id":"2","o":"21"]] 
def m2 = [["o":"11","t":"t1"],["o":"11","t":"t2"],["o":"21","t":"t1"]] 

我想导致

[["id":"1","t":"t1"],["id":"1","t":"t2"],["id":"2","t":"t1"]] 

我目前的迭代地图和做到这一点。我正在寻找一个解决方案，使用Gpath和findAll

谢谢， Sreehari。

Answer 1

您可以transpose两份名单，并从每个列表中的条目（id或t）：

def fn = { m1, m2 -> 
    return [m1,m2] 
     .transpose() 
     .collect { [ id: it.first().id, t: it.last().t ] } 
} 


def m1 = [["id":"1","o":"11"],["id":"1","o":"12"],["id":"2","o":"21"]] 
def m2 = [["o":"11","t":"t1"],["o":"11","t":"t2"],["o":"21","t":"t1"]] 


assert fn(m1, m2) == 
     [["id":"1","t":"t1"],["id":"1","t":"t2"],["id":"2","t":"t1"]] 

Answer 2

可以使用转置到地图拉链成对，然后通过地图键结合对和过滤器：

[m1, m2] 
    .transpose() 
    .collect { (it[0] + it[1]).subMap(['id', 't']) } 

计算结果为

[[id:1, t:t1], [id:1, t:t2], [id:2, t:t1]] 

这适用于groovysh使用Groovy-2.4 .4，与jdk7或jdk8。